A solid compound `XY` has `NaCl` structure. If the radius of the cation is `100` pm, the radius of the anion `(Y^(-))` will be

Options

A. 275.1
B. 322.5 pm
C. 241.5 pm
D. 165.7 pm

Answer

C

Explanation

NaCl has face-centred cubic arrangement of `Cl^-` ions and `Na^+` ions are present in the octahedral voids. Hence, for such a solid radius of cation =0.414 x radius of the anion ( or `r_+/r_(-)`=0.414)
i.e., `r_+= 0.414xxr_(-)`
or `r_(-)=r_+/0.414=100/0.414`=241.5 pm